For any real numbers a,b with a < b, let Sa,b(x)
denote a selection function, defined as

Using this function we can express the integral of a
function f(x) over the range from x = a to b as follows

Thus, instead of specifying the range of interest by
imposing explicit limits on the integration, we can formally integrate over
all real values of x, using the multiplier Sa,b(x) to select the
desired range. The total integrand equals 0 outside the specified range, and
f(x) inside the specified range. The average value of the function f(x) from
x = a to b is given by dividing the above integral by the length of the
interval, i.e.,

If the values of a and b are extremely close together, the
length of the interval approaches zero, and the average value of the function
over that interval approaches f(a). Putting b = a + e, notice that we can always set a = 0 by simply shifting the
argument of the selection function by a, so we have

It’s often convenient to consider this scaled selection
function in the limit as e goes to
zero. In this limit we have

This leads us to define the so-called Dirac delta “function”
as follows

Strictly speaking this isn’t an actual function, because
it is zero everywhere except at x = 0, where it is infinite. However, the
symbol d(x) may be regarded as useful
shorthand for writing certain limiting cases of integrals. By definition, the
integral of d(x) over any range
containing x = 0 is equal to 1, and the integral of d(x) over any range not containing x = 0 is equal to 0. Also,
the delta “function” has a well-defined effect when it appears in the
integrand of any integral. For example, we can write equation (1) without
explicitly referring to the limiting operation as

It’s often useful to be able to form the composition of
the delta function with some other function. In general, consider d(g(x)) where g(x) is an ordinary function
with n distinct roots x1, x2, …, xn. This
means that g(x) = 0 for each of these n values of x, so these are the places
where Se(g(x))/e contains a spike, as depicted below for a
function with n = 4 roots.

Notice that the widths of the spikes are not all the same
in this view, even though this is drawn for a single fixed value of e. The reason is that the argument of S is
g(x), whereas we have plotted against x. The e
limit for the selection function is an increment of g rather than of x. Increments
of g are related to increments of x by the derivative dg/dx = g’, so we have
dg = g’dx. Therefore, if we were to replace the argument g(x) near the root x1
(for example) by the function x – x1, we would get a spike at the
same location, but for a given e it
would be too narrow by the factor |g’(x1)|. (We take the absolute
value, because the ratio of widths just depends on the magnitude of the
slope, not the sign.) To give the right integrated area, we must divide the
height by the same factor.

For example, consider just one of the roots, say x1,
and factor g(x) into the form g(x) = (Ax + B)h(x) where x1 = -B/A. To represent the contribution of this
spike to the total delta function, we can replace the argument g(x) by (x +
B/A), because they both have a root at x1. Of course, x + B/A has
the same slope as x, so our previous correction factor must be applied. Thus
we have

In the limit as e
goes to zero, an expression of this form gives the contribution to the delta
function of each root, so we have the relation

Returning to the basic definition of the delta function,
we note that it’s possible to define d(x)
in several equivalent ways, including as the limit of certain continuous
functions. One particularly useful expression is based on the Cauchy
distribution

To show explicitly that equation (2) is valid using this
simple algebraic definition of d(x),
note that any function g(x) with n distinct real roots x1 to xn
can be written in the form

where h(x) has no real roots. (See below for comments on
the restriction to real roots.) We wish to show that

For any given e we
need to show that the summation, with the delta functions written explicitly
in the form of (3), equals the original expression with some (possibly
re-scaled) value of e. We have

Consolidating the right-hand summation, and neglecting
powers of e higher than the first in the numerator, and higher than second in
the denominator, we get

Setting x equal to any one of the roots xk,
most of the terms vanish, and we are left with

Noting that

our expression reduces to

Multiplying through the numerator and denominator by h(xk)2,
we arrive at

The right hand expression (multiplied by 1/p) is indeed of the required form, with an
epsilon value that is just a constant, g’(xk), times the epsilon
value of the individual delta functions in the summation. This confirms that
(2) is valid for the delta function as defined by (3).

To illustrate the delicacy of the delta functions in
applications, consider how we might compute the length of a plane locus by
integrating, over the entire plane, the delta function of an expression that
vanishes on that locus. For example, one might think that the perimeter of a
circle of radius R could be computed as

However, taking just the inner integral first, for any
constant x in the range –R to +R, we have

Inserting this back into the double integral (4), and
restricting the integration on x to the range –R to +R (which is permissible,
since we know the delta function doesn’t vanish outside that range) we get

The result is a constant, p,
regardless of the radius R, so this clearly does not represent the
circumference of the circle. To confirm this result, we might convert to
polar coordinates r,q such that x = r
cos(q) and y = r sin(q). As described in the note on change of variables in multiple
integrals, we have dy dx = r dr dq,
so the double integral (4) becomes

Notice that the argument of the second delta function on
the right side is zero only when r equals –R, whereas we are integrating over
the range from r = 0 to +R, so that term has no contribution. Thus we have

This confirms that the integral is simply p, independent of R, so it does not represent
the circumference, even though the argument of the delta function vanishes
precisely on the perimeter of the circle. What has gone wrong? The problem is
that we took a quadratic function as the argument of the delta function,
whereas we ought to take a linear function. For example, in polar coordinates
the perimeter of the circle corresponds to the points where r – R vanishes,
not where r + R vanishes, so we ought to write

which is indeed the circumference of the circle. Likewise
in Cartesian coordinates we should just take the relevant linear condition
for the argument of the delta function, and write

Focusing again on just the inner integral for any constant
x in the range from –R to R, the argument of the delta function is

which has the derivative

The two roots of g(y) are

so we have

Therefore, making use of equation (2), the inner integral
is

Inserting this back into the double integral (5), and
again restricting the integration to the range –R to +R as explained
previously, we get

as expected. This shows the importance of choosing the
right argument for the delta function.

We mentioned previously that the roots of g(x) appearing
in equation (2) are real, which is to say, we exclude any roots that are not
on the real line. This is because we are integrating on the real numbers, so
any roots off the real number line will not be encountered, and hence don’t
contribute to the integration. However, we could conceivably apply something
like a delta function to integrations over the entire complex plane, in which
case the complex roots of the argument would be significant. Of course, we
could simply use a product of two ordinary one-dimensional delta functions,
one for the real part and one for the imaginary part of the argument, but
there might also be a more natural generalization of the delta function to
complex numbers.

One of the main uses of the delta function arises from its
properties when subjected to the common integral transforms. For example, by
definition, the Laplace transform of the delta function is

To show how this is useful, consider an ordinary differential
equation

where f(t) = 0 for all t < 0. The driving function f(t)
may be of a form that is difficult to handle, but if we replaced f(t) by an
impulse function d(t – t’) occurring
at some time t’, we can easily solve the equation using Laplace transforms.
Let this solution for an impulse at time t’ be denoted by G(t,t’), so we have

(The function G is a simple example of what is called a
Green’s function.) Now, notice that the original driving function can be
expressed as

Bringing the differential operator outside the integral,
we have

and hence the solution x(t) of the original problem is the
second factor on the right side, i.e., the solution can be written in terms
of the impulse solution G(t,t’) as

Another reason for the usefulness of Dirac’s delta
function is that it has the simple Fourier transform

Applying the inverse Fourier transform, we might think the
delta function itself can be expressed as

However, this integral diverges, so this is not a
well-defined expression. Nevertheless, if we cut off the integration at large
but finite limits, we do get a function that converges on the delta function
as the limits increase, so this expression is often presented as a formally
valid representation of the delta function.

Incidentally, several features of the delta function
discussed above can be seen in Feynman’s 1949 paper on quantum
electrodynamics (for which he was later awarded the Nobel prize). His
objective is to derive an expression for the amplitude of a particular
interaction between two charged particles, corresponding to the Coulomb
potential e2/r where e is the electric charge on each particle and
r is the distance between the particles. In his notation, the interaction is
“turned on” when one particle is at a time and place denoted by event 5, and
the other is at the time and place denoted by event 6. He begins with a
double integral over the path parameters dt5
and dt6 , and to accept
contributions only when t5 = t6 he multiplies the
integrand by the delta function d(t5
– t6), treating the Coulomb interaction non-relativistically as if
it acted instantaneously. Thus his expression included a factor of the form

But then he notes that, relativistically, the interaction
is not instantaneous, but is retarded by the light-speed delay in propagating
from one particle to the other. Thus, is we define the symbols t56
= t5 – t6 and r56 = r5 – r6,
Feynman proposes to modify the formula, substituting in place of d(t56) the delta function d(t56 – r56), which
signifies that the integral will accept contributions when the particle at 5
is on the forward light cone of the particle at 6. (Note that we are using
units such that the speed of light has the value 1.) But now he says “this
turns out to be not quite right, for when this interaction is represented by
photons, they must be of only positive energy, while the Fourier transform of
d(t56 – r56)
contains frequencies of both signs”. He is referring to the representation of
the delta function given by equation (6), noting that the integral extends
over values of the frequency parameter w
from negative to positive infinity. (Oddly enough, he doesn’t mention that
this representation doesn’t actually converge.) To remedy this, he defines a
new type of delta function, denoted by d+(x),
consisting of only the positive frequency parts of the usual delta function,
i.e.,

Here the integral ranges from 0 to positive infinite, and
we have multiplied by 2 to normalize the result so that the integral of d+(x) over all x is 1. Also,
Feynman wants to account for the reverse interaction, when the particle at 6
is on the forward light cone of the particle at 5, which happens when t56
+ r56 vanishes. He says we need to average the two possibilities,
so he replaces (7) with

But recall from our discussion of the composition of
functions that

Making use of this fact (which applies to d+ as well as to d), Feynman arrives are the
relativistically invariant factor

Without knowing the background for this expression, one
might wonder how the amplitude for the Coulomb interaction e2/r
can be independent of the distance r56, but we see that the
inverse distance is implicit in the delta function of the relativistic
squared interval.